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        riddles >> general problem-solving / chatting / whatever >> Help with antiderivative
        
(Message started by: Hooie on May 7^th, 2004, 4:49pm)

Title: Help with antiderivative
Post by Hooie on May 7^th, 2004, 4:49pm

Does arctan (e^x) have an antiderivative?

I tried doing it by parts, but once you do it twice, you end up back where you started.

I tried rewriting it as (arctan x)/x and then using parts, but the same thing happens.

:-\ I'm stuck! Can someone help?

Title: Re: Help with antiderivative
Post by Sir Col on May 9^th, 2004, 11:31am

I find the QuickMath.com (http://quickmath.com)'s online verison of Mathematica handy for these types of problems. Although I have no idea how it did it, you can click here (http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=basic&expression=arctan(e^x)&variables=x&showReply=true&prefInput=standard&prefNames=single&fontSize=20&style=traditional) for your rather ugly integral.

Title: Re: Help with antiderivative
Post by THUDandBLUNDER on May 9^th, 2004, 3:02pm

I wonder why it seems to integrate [smiley=smallint.gif](sinx)^0.5.dx (http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=basic&expression=(sin(x))^(0.5)
&variables=x&showReply=true&prefInput=standard&prefNames=single&fontSize=20&style=traditional) but not [smiley=smallint.gif](sinx)^1/2.dx (http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=basic&expression=(sin(x))^(1/2)
&variables=x&showReply=true&prefInput=standard&prefNames=single&fontSize=20&style=traditional)

Well, for (sinx)^1/2 it sometimes gives no answer (output = input), and sometimes gives an answer of
'-2E[1/2{(pi/2) - x} | 2]', whatever that is.

Nor do I understand the answer for (sinx)^0.5.

Title: Re: Help with antiderivative
Post by Icarus on May 10^th, 2004, 5:37pm

First, for the original integral. The reason QuickMath was able to integrate it with the closed form

x arctan(e^x) + (ix/2)(ln(1+ie^x) - ln(1-ie^x)) + (i/2)(Li₂(-ie^x) - Li₂(-ie^x))

is because it has a larger stock of "elementary functions" than you do. Li₂(x) is the "dilogarithm": [sum]_k=1^[supinfty] x^k/k².

Concerning [int](sin x)^1/2 dx, I always get a response, though if I don't phrase it right, the thing interprets "s", "i", "n" as constants rather than the name of a function. :P

If you use the advanced feature, and select "text" output, you can get clues as to what the functions are. For sin(x)^1/2, the function is called "elliptice", which apparently means it is an elliptic integral of the second kind, though the notation I would have expected is -2 E(2, [pi]/4 - x/2). The Elliptic integral function of the second kind is defined by

E(k, x) = [int]₀^x (1 - k²sin²(t))^1/2 dt.

(An exponent of -1/2 is an elliptic integral of the first kind: F(k,x). There is also a more complicated third kind).

The ₂F₁ function is called "hypergeometric2F1". Here is what Mathworld has to say about it. (http://mathworld.wolfram.com/GeneralizedHypergeometricFunction.html)

The problem with using these results is that unless you are intimately familiar with these functions, these anti-derivatives tell you nothing.

As a side-note, all continuous (and many discontinuous) functions have antiderivatives. However only a few of them have nice closed form expressions. You can define an antiderivative for arctan(e^x) by the equation f(x) = [int]₀^x arctan(e^t) dt. From a calculational standpoint (easy in calculating the values of f(x) for given x), this might be a better formula than the dilogarithm equation above. We have some fairly stable means of estimating integrals. The calculations might actually prove easier by that means than by calculating the dilogarithm (or perhaps not - the dilogarithm series is fairly simple too). But I would almost guarantee it easier to numerically integrate sin(x)^0.5 than to calculate the value from the hypergeometric function.

Title: Re: Help with antiderivative
Post by SWF on May 11^th, 2004, 5:55pm

Sometimes it is easier to just expand in a series yourself instead of messing around with those less common functions. In this case, the anti-derivative, F(x), turns out to be an easy to use expression (sums are taken over n= 0 to [infty]):

(for x<=0) F(x)= C + [sum](-1)ⁿ exp((2*n+1)x) / (2*n+1)²

(for x>0) F(x)= C + [pi]x/2 + [sum](-1)ⁿ exp(-(2*n+1)x) / (2*n+1)²

C is that arbitrary constant math teachers love to dock you points for forgetting. Of course, integral between any given limits a and b is F(b)-F(a), and the value of C does not matter.

Note that integral of atan(exp(x)) from -2 to +2 is [pi].